We will rely freely on the fact that if an additive and multiplicative map \(\sigma \colon \mathbb {C}\rightarrow \mathbb {C}\)  is bounded from above on a set \(\Gamma \) , then it remains bounded from above on a set \(\Gamma _1\)  obtained from \(\Gamma \)  by applying any of the following four procedures finitely many times:
(i) taking a subset, i.e., \(\Gamma _1\subseteq \Gamma \) , (ii) translating by \(\mu \in \mathbb {C}\) , i.e., \(\Gamma _1=\mu +\Gamma \) , (iii) rotating/expanding, i.e., \(\Gamma _1=\mu \Gamma \) , and (iv) taking the union of finitely many sets obtained in previous iterations.
Let \({\bf r}\colon [0,1]\rightarrow \Gamma \)  be the corresponding non-constant path. By restricting and parameterizing the domain we can assume that \({\bf r}(0)\ne {\bf r}(1)\) . By translating and rotating/expanding the resulting curve we can further assume that \({\bf r}(0)=0\)  and \({\bf r}(1)=1\) .
If the curve \(\Gamma \)  contains a line segment then we can further translate and rotate/expand it to assume that this line segment equals the interval \([0,1]\) . Then \(\sigma \)  is bounded on the closed unit disk because if \(a,b\in [0,1]\) , then \(\sigma (a+ib) =\sigma (a) + \sigma (i)\sigma (b)\) . As such, \(\sigma \)  is continuous (see, e.g., [1]}) and the result follows.
Assume \(\Gamma \)  contains no line segment. Being compact, it is contained in the smallest solid rectangle \(Q\)  whose sides are parallel to real/imaginary axis. Now, the union \(\Gamma \cup (1+\Gamma )\)  is still a curve because the end point of a path \({\bf r}\)  in \(\Gamma \)  coincides with the initial point of a translated path \(t\mapsto 1+{\bf r}(t)\)  of \((1+\Gamma )\) . By induction we see that the finite union
\(\bigcup _{k=0}^{n-1}(k+\Gamma )\)  which, by the abuse of notation we again denote by \(\Gamma \) , is still a curve which connects 0 to \(n\) ; here an integer \(n\)  is chosen to be bigger than four-times the height of a rectangle \(Q\) . By another abuse of notation we denote the union of translated solid rectangles \(\bigcup _{k=0}^{n-1} (k+Q)\)  again by \(Q\) .
We now rotate \(\Gamma \)  by \(\pi /2\) , i.e., we consider \(i\Gamma \) , and translate \(\Gamma \)  as well as its rotated version. The resulting union (with negative sign in the exponent denoting that the parametrization of \(\Gamma \)  must be reversed),
\(\hat{\Gamma }:=\Gamma ^-\cup (i\Gamma )\cup (ni+\Gamma )\cup (n+i\Gamma ^-)\) 
is a closed curve, which connects in clock-wise direction 0 to \(ni\) , to \(n(i+1)\)  to \(n\)  and back to 0. Also, this closed curve is contained in the union of four solid rectangles
\(\hat{Q}:=Q\cup (i Q)\cup (ni+Q)\cup (n+iQ)\) 
each obtained by rotating \(Q\)  by \(\pi /2\)  and/or translating it. Since \(n\)  is at least four times the height of \(Q\) , we see that the above union properly encircles, but does not contain a point \(\alpha :=\frac{n}{2}(1+i)\) .
<FIGURE>Notice that a clock-wise oriented rectangle \(\hat{R}:=[0,n]\cup i[0,n]\cup (ni+[0,n])\cup (n+i[0,n])\subseteq \hat{Q}\) , i.e., the closed polygonal curve which connects \(0 \rightarrow (in)\rightarrow n(1+i)\rightarrow n\rightarrow 0\)  is a deformation of \(\hat{Q}\)  within \(\hat{Q}\)  (that is, the identity map on \(\hat{Q}\)  is homotopic, within \(\hat{Q}\) , to a map from \(\hat{Q}\)  onto \(\hat{R}\) ). This deformation then also induces a deformation of \(\hat{\Gamma }\)  onto \(\hat{R}\)  within \(\hat{Q}\)  which never passes through \(\alpha :=\frac{n}{2}(1+i)\)  nor through \(\infty \) . Since \(\hat{R}\)  separates \(\mathbb {C}\)  by the Jordan Curve Theorem, the same holds for \(\hat{\Gamma }\)  by [2]}, and in fact \(\alpha \)  is in a bounded component of \(\mathbb {C}\setminus \hat{\Gamma }\) .
Since \(\sigma \)  is additive, it is clearly bounded from above also on the subset \(\hat{\Gamma }-\hat{\Gamma }\subseteq \mathbb {C}\) . We claim that this subset has a nonempty interior, so it must also contain a line segment, and hence, as shown above, \(\sigma \)  is continuous. To see that \(\hat{\Gamma }-\hat{\Gamma }\subseteq \mathbb {C}\)  contains interior points, we borrow the idea from a paper by Simon and Taylor [3]}.
Let \(C_\infty \)  be the unbounded component of \(\mathbb {C}\setminus \hat{\Gamma }\)  and let \(C_0:=\mathbb {C}\setminus (C_\infty \cup \hat{\Gamma })\)  be the union of its bounded components. Notice that \(C_\infty \)  is an open set whose boundary is contained in \(\hat{\Gamma }\)  (see, e.g., [2]}),
so that \((C_\infty \cup \hat{\Gamma })=\overline{C_\infty }\cup \hat{\Gamma }\)  is closed and as such its complement, \(C_0\)  is open.
Let
\(\Omega :=\lbrace x\in \mathbb {C};\;\;\exists \,a_1,a_2\in \hat{\Gamma }: a_1\in x+C_0\,\hbox{ and }\,a_2\in x+C_\infty \rbrace .\) 
This is clearly an open subset. It is also nonempty: in fact, let \(\alpha \in C_0\)  be a point in the bounded component. Consider the two horizontal rays emanating from \(\alpha \) , that is, \(L^{-}:=(-\infty ,0]+\alpha \)  and \(L^{+}:=\alpha +[0,\infty ) \) . Since \(\hat{\Gamma }\)  separates \(\mathbb {C}\) , both rays must intersect it. Let \(a_1\in \hat{\Gamma }\cap L^{-}\)  be the closest point to \(\alpha \)  on the left ray which also belongs to a compact closed curve \(\hat{\Gamma }\)  and let \(a_2\in \hat{\Gamma }\cap L^+\)  be the furthest point from \(\alpha \)  on the right ray which still belongs to \(\hat{\Gamma }\) . By moving \(\hat{\Gamma }\)  slightly to the left, i.e., by adding small enough real \(x<0\)  to \(\hat{\Gamma }\)  we achieve that \(a_1\in \hat{\Gamma }\)  belongs to the bounded component of \(\mathbb {C}\setminus (x +\hat{\Gamma })\) , that is, \(a_1\in x+C_0\)  while \(a_2\in \hat{\Gamma }\)  belongs to its unbounded component, that is, \(a_2\in x+C_\infty \) . Indeed, \(\Omega \)  is nonempty.
Assume there is \(x\in \Omega \)  such that \(x\notin \hat{\Gamma }-\hat{\Gamma }\) . Then \((x+\hat{\Gamma })\cap \hat{\Gamma } = \varnothing \) . It follows that a connected set \(-x+\hat{\Gamma }\)  is completely contained in the union of disjoint open subsets \(C_\infty \cup C_0\)  (namely, \(-x+\hat{\Gamma }\subseteq \mathbb {C}= C_0\sqcup \hat{\Gamma }\sqcup C_\infty \) , the disjoint union), so \(\hat{\Gamma }\subseteq (x+C_0)\sqcup (x+C_\infty )\) . Also, by definition of \(\Omega \) , both open sets \((x+C_0)\)  and \((x+C_\infty )\)  contain a portion of \(\hat{\Gamma }\) , so they partition a connected set \(\hat{\Gamma }\) , a contradiction. Indeed, \(\Omega \)  is an open subset contained in \(\hat{\Gamma }-\hat{\Gamma }\) .
